On 2007-02-12,
[email protected] <
[email protected]> wrote:
> Ben C? writes:
[...]
>> If the caliper is 1/4 the radius of the tyre, and situated at 3
>> o'clock, then the axle force will be only 14 degrees from vertical,
>> not 45. This makes vertical dropouts and a 3 o'clock caliper much
>> riskier than I had thought.
>
>> So yes, not just a large magnitude but a direction close enough to
>> the dropout exit direction that I can believe the axle could roll
>> down the dropout.
>
> I don't know with what values you are starting but the limiting
> condition is when the rear wheel lifts from braking.
That's my basic model, with a friction coefficient of 1.0, and so the
braking force is horizontal and rearward.
> Rough terrain is not required for this and only clouds the issue.
It does cloud the issue, but it needs to be considered in case it is a
factor.
> Smooth dry pavement will do. At liftoff the line of action of the
> rider/bicycle CG can act at a 45° angle to the tire contact (which is
> not the angle of action at the dropouts).
What difference does the line of action of the rider/bicycle CG make?
I'm just considering the forces applied to the disk by the caliper and
to the tyre by the road (contact force, vertically upwards, and braking
force, normal to that, horizontally rearwards).
> If the disk is roughly 1/4 the wheel diameter, a downward force about
> 8x the upward force occurs on that end of the axle, perpendicular to
> the brake pad to axle axis. The 8x effect arises from the brake force
> being concentrated on one dropout and the wheel load resting on two.
Yes, good point.
> The wheel does not lift out of the dropouts but is canted with one end
> shifting downward in one dropout only to cause wheel lockup.
If the wheel actually did eject, I imagine the bike would sort of hop up
over the wheel.
Anyway, I now revise my calculations again
This time I get 24 degrees from vertical for Ben Micklem's wheel
(caliper at 2:30), which makes it 44 degrees between ejection force and
dropout direction (his dropouts are 20 degrees forwards). I get only 8
degrees from vertical for a caliper at 3 o'clock.
Here is my working in the form of a GNU Octave script if anyone has the
energy to check it.
function phi = ejection_angle(theta, R, r)
% theta is angle of caliper in radians, measured anticlockwise with 0 at 3
% o'clock. R is tyre radius, r is disk radius. phi is angle of ejection
% force from vertical in degrees.
% The caliper force. x +'ve is rearward, y -'ve is downward.
c = [sin(theta), -cos(theta)];
% Its magnitude is scaled by difference in radius between disk and tyre,
% and multiplied again by 2 since the disk is only on one side of the axle.
c *= 2*R / r;
% Add the braking force and the contact force (friction coefficient is 1.0)
% to give the net force f on the wheel under braking.
f = c + [1, 0] + [0, 1];
% Work out that vector's direction from vertical
phi = atan(f(1) / f(2));
phi *= -360 / (2*pi);
endfunction
% Caliper at 2:30.
theta = (0.5 / 12) * 2*pi;
disp(ejection_angle(theta, 675.14, 160));
% Caliper at 3:00
disp(ejection_angle(0, 675.14, 160));
